# Moment of inertia of thin circular hoop

Why does a thin circular hoop of radius $r$ and mass $m$ have the following moments of inertia?

$I_x=I_y=\frac{mr^2}{2}$ and the sum is $I_z=I_x+I_y$?

The formula of finding moment of inertia is:

$I = \int r^2dm$, where $dm=\rho dV=2r\pi \rho$.

How to obtain $I_x=I_y=\frac{mr^2}{2}$ from here onwards?

A thin circular hoop of radius $r$ centred at $(0,0,0)$ and contained in the $xy$-plane, $$\{(x,y,z)\in\mathbb{R}^3: x^2+y^2=r^2, z=0\}$$
is a one-dimensional solid and the moments of inertia with respect to the $x$-axis ($y=z=0$), $y$-axis ($x=z=0$), and $z$-axis ($y=x=0$) are: $$I_x=\int_{\theta=0}^{2\pi}y^2\cdot (\delta r d\theta)\ ,\ I_y=\int_{\theta=0}^{2\pi}x^2\cdot (\delta r d\theta) \ \mbox{and}\ I_z=\int_{\theta=0}^{2\pi}(x^2+y^2)\cdot (\delta r d\theta)=I_x+I_y$$ where $x=r\cos\theta$, $y=r\sin\theta$ and $\delta$ is the linear density.

Here we assume that $\delta$ is constant and therefore $m=2\pi r \delta$.

Can you take it from here?

• First of all thank you very much. If I then evaluate the integral I get $I_x=2pi*r*delta *y^2=my^2$ but does that mean that I evaluated integral wrong (if yes then sorry) or that my initial answer that it should be divided by 2 wrong? – user203 Sep 26 '16 at 10:17
• @user203 $y(\theta)=r\sin(\theta)$ It dipends on $\theta$! See my edited answer and let me know. – Robert Z Sep 26 '16 at 10:21
• Amazing explanation ! I used $sin(theta)^2=1/2-1/2*cos(2theta)$ and now I got it. Thank you very much – user203 Sep 26 '16 at 12:43
• @user203 Well done. – Robert Z Sep 26 '16 at 12:51

$I_z=mr^2$ because all of the mass is situated at distance $r$ from the axis. By symmetry $I_x=I_y$. Generally $I_z=I_x+I_y$. It follows that $I_x=I_y=\frac12mr^2$.